ANALOG INTEGRATED

CIRCUITS

(TWO STAGE AMPLIFIER)

ECE 196 LAB 10

MENCHIE LABADAN

BS ECE IV

Submitted to:

PROF. OLGA LABAJO GERASTA

EXPERIMENT STEPSEXPERIMENT STEPSEXPERIMENT STEPSEXPERIMENT STEPS

TWO STAGE OP AMPTWO STAGE OP AMPTWO STAGE OP AMPTWO STAGE OP AMP

HSPICE CODE:

MOSFETS:

GAIN

GRAPH

DIFFERENTIAL OP AMPDIFFERENTIAL OP AMPDIFFERENTIAL OP AMPDIFFERENTIAL OP AMP

Mosfets mn6 and mn7 are removed. The new node of vout is at node x. The capacitor is now

connected at node x.

MOSFETS:

GRAPH

PERFORMANCE COMPARISON

Structure gain (dB) -3dB (MHz) G.B (MHz)

two stage amplifier 56.9 6.9734 7.2623

differential amplifier 43 0.48794 0.53044

Hence from the simulation results, the performance of the Two-Stage Operational amplifier is much

better than Differential Amplifier.

We have the following HPICE Codes

All the mosfets operate in saturation region as seen below and Id does not exceed 10uA,

The figure shows the graph of the phase margin and the gain.

The graph below shows the Vin Vout curve expressed in terms of time,

Hence the Phase Margin is,

PM = 180 – 177.86 = 2.14

As we connected the circuit as a unit gain buffer, we can see that the phase margin is less than 0

degrees. Hence we add the negative value of phase margin to 180 degrees to get our final phase

margin, which is 2.14. Hence the system is unstable. It is because the typical value of phase margin

must be larger than 45 degrees for it to be considered a stable system.

In my circuit, I added a compensation capacitor of 2.15pF. The HPICE code is,

Phase Margin = 180 – 135.07 = 44.93° ≈ 45°

After compensation, pole1 frequency decreased and pole 2 frequency increased making the phase

margin around 45° and also the system becomes more stable.

Effect on Phase Margin

Hspice code

at 2.15p PM=180-136.47=43.53

at 3pf PM=180-128.87=51.53

at 5pf PM=180-120.27=59.53

at 10pf PM= 180- 111.7=68.3

Conclusion: As we increase Cc the phase margin also increased

Effect on Slew Rate

Hspice code

At 2pF

At 3pF

At 5pF

At 10pF

Conclusion:As we increase Cc the slew rate decreases, because larger Cc need more time to charge

or discharge

Rc = 1/gm6 = 1/138.2344u = 7.234kΩ

First, we choose Rc =1.234k ohm to move the zero to infinite.

From the simulation, we can see that the gain does not increase after second pole, it means that the

zero is moved to infinite. And PM = 180-127.2= 52.8

We try a new value of Rc which is,

Rc =( 1/gm6)((Cc+Cl)/Cc) = (1/138.2344u) (2.15p+10p)/(2.15p) = 40.88kΩ

At Rc= 40.88kΩ

Phase Margin = 180 – 93.19 = 86.81

Hence the phase margin increased.

Input Common Range (ICMR) Typical Value – 0.7 to 3V

Vin and Vout are shorted.

The graph is,

ICMR = 1.77 - 0.572= 1.198. Hence it is still is in the range of the typical value for ICMR.

Output Voltage Swing

We have the hspice code,

The graph below shows the Vin and Vout curve for the output voltage swing,

Where Vout = -(100R/10R)Vin = -10Vin

R = 1 MΩ

COMMON MODE REJECTION RATIO

Common Mode Rejection Ratio (CMRR): the ability of ejecting

common mode signal.

HSPICE CODE:

two stage amp-CMRR_mench

.lib 'C:\synopsys\rf018.l' TT

.option post

mp3 a a vdd vdd pch l=1u w=5.5u

mp4 x a vdd vdd pch l=1u w=5.5u

mn1 x vin+ b gnd nch l=1u w=2.5u

mn2 a vin- b gnd nch l=1u w=2.5u

mn5 b c gnd gnd nch l=1u w=5u

mn8 c c gnd gnd nch l=1u w=5u

mp6 vout x vdd vdd pch l=1u w=33u

mn7 vout c gnd gnd nch l=1u w=12u

vdd vdd gnd 1.8

Cl vout gnd 10p

Iref vdd c 3.7u

vin+ vin+ gnd dc 0.9 ac 1

vin- vin- gnd dc 0.9 ac 0

.ac dec 10 100 10g

.print vdb(vout)

.alter

vin- vin- gnd dc 0.9 ac 1

.op

.end

CMRR = 88.3

βSLEW RATE AND SETTLING TIME

Slew Rate: the maximum rate of change dvo/dt is called slew

rate.

Settling Time: the time during output signal be stable shown as

left figure

We have the following HSPICE CODE for Vinp which has a peak voltage of 1V

The transient time and the slew rate is,

We have the following graph,

For a Vip of 1.4 V we have the following hspice code :

The transient time and slew rate is,

The graph is,

From the simulation, we can see that the slew rate will increase while the swing of input pulse

increases.

QUESTIONQUESTIONQUESTIONQUESTION

Ans

The common source is used for higher gain. If we use other structures, there’s a posiibility that it the

gain would be much lower.